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Chemistry Marathon (HSC) (2 Viewers)

deswa1

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This is for people doing industrial chem- I won't post any more optional questions because not everyone can do them obviously but this one is fantastic and deserves to posted:

The ionisation of acetic acid can be written as CH3COOH -> <- CH3COO (-) + H (+). The equilibrium constant for this reaction is 1.8x10^-5. A 0.5mol/L solution of acetic acid is made up. Calculate the pH of the resultant solution at equilbrium.
 

barbernator

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This is for people doing industrial chem- I won't post any more optional questions because not everyone can do them obviously but this one is fantastic and deserves to posted:

The ionisation of acetic acid can be written as CH3COOH -> <- CH3COO (-) + H (+). The equilibrium constant for this reaction is 1.8x10^-5. A 0.5mol/L solution of acetic acid is made up. Calculate the pH of the resultant solution at equilbrium.
pH = 12.7 ??
 

Sanjeet

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This is for people doing industrial chem- I won't post any more optional questions because not everyone can do them obviously but this one is fantastic and deserves to posted:

The ionisation of acetic acid can be written as CH3COOH -> <- CH3COO (-) + H (+). The equilibrium constant for this reaction is 1.8x10^-5. A 0.5mol/L solution of acetic acid is made up. Calculate the pH of the resultant solution at equilbrium.
Anyone have a solution?
 

barbernator

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Anyone have a solution?
1. Write out the equilibrium expression
2. We want to replace everything that is not [H+] with either [H+] or a constant
3. We know from standard ionisation that [H+] = [CH3COO-]
4. as 0.5 mol/L of Acetic acid is originally placed in the beaker, and these molecules either stay in their molecular form or ionise, we can say that [H+] + [CH3COOH] = 0.5
5. Replace all things other than [H+]
6. Solve quadratic taking the positive solution
7. Use pH = -log[H+]
8. Presto!
 
Last edited:

someth1ng

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1. Write out the equilibrium expression
2. We want to replace everything that is not [H+] with either [H+] or a constant
3. We know from standard ionisation that [H+] = [CH3COO-]
4. as 0.5 mol/L of Acetic acid is originally placed in the beaker, and these molecules either stay in their molecular form or ionise, we can say that [H+] + [CH3COOH] = 0.5
5. Replace all things other than [H+]
6. Solve quadratic taking the positive solution
7. Use pH = -log[H+]
8. Presto!
lol...fuck that - won't be in the exam :p
 

madharris

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Is industrial chem good?
Shipwrecks is so boring imo
 

cerebellum

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I also got pH = 2.52 (3 sig figs).
What else you got, bring it.
:)
 

madharris

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A student said that the use of ethanol as a fuel is entirely carbon neutral. Evaluate this statement with appropriate equations
Not sure how many marks but oh well, I'll try :)

Ethanol is a renewable resource which can be made from cellulose. Cellulose is a naturally occuring condensation polymer which makes up about 50% of the earths biomassIt is manufactured by photosynthesis in plants, fermentation then combustion in vehicles. THis means that it's 'carbon neutral':
photosynthesis: 6CO2(g) + 6H2(l)L --> 6O2(g) + C6H12O6(aq)
fermentation: C6H12O6(g) --yeast--> 2C2H5OH(aq) + 2CO2(g)
Combustion: C2H5OH(aq) + 3O2(g) --> 2CO2(g) + 3H2O(g)
The amount of CO2 released in fermentation and combustion is equimolar (not sure of this is the right word) to the CO2 absorbed in photosynthesis
However the energy in harvesting, manufacturing, transporting and disposal of the products generally comes from and gives CO2.
In the fermentation process we also only gain 15% ethanol, meaning more energy is used in the process, creating more CO2
Overall ethanol is not a carbon neutral fuel and therefore is not an energy source effective as a countermeasure to climate change. More research is needed in order to to be used as a alternative fuel source which is entirely carbon neutral.

Anyone willing to correct/add to that go ahead :)

Outline the changes that have occured in the atmospheric ozone concentrations above Antarctica during the past 20 years and explain how this information was obtained. - 4 marks
 

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